A complex (phasor) representation is implied, so taking real parts when all is said and done is required of the fields. For the energy momentum tensor the Geometric Algebra form, modified for complex fields, is used

The assumed four vector potential will be written

Subject to the requirement that is a solution of Maxwell’s equation

To avoid latex hell, no special notation will be used for the Fourier coefficients,

When convenient and unambiguous, this dependence will be implied.

Having picked a time and space representation for the field, it will be natural to express both the four potential and the gradient as scalar plus spatial vector, instead of using the Dirac basis. For the gradient this is

and for the four potential (or the Fourier transform functions), this is

Setup

The field bivector is required for the energy momentum tensor. This is

This last term is a spatial curl and the field is then

Applied to the Fourier representation this is

It is only the real parts of this that we are actually interested in, unless physical meaning can be assigned to the complete complex vector field.

Constraints supplied by Maxwell’s equation.

A Fourier transform solution of Maxwell’s vacuum equation has been assumed. Having expressed the Faraday bivector in terms of spatial vector quantities, it is more convenient to do this back substitution into after pre-multiplying Maxwell’s equation by , namely

Applied to the spatially decomposed field as specified in (2.7), this is

All grades of this equation must simultaneously equal zero, and the bivector grades have canceled (assuming commuting space and time partials), leaving two equations of constraint for the system

It is immediately evident that a gauge transformation could be immediately helpful to simplify things. In [3] the gauge choice is used. From (3.11) this implies that . Bohm argues that for this current and charge free case this implies , but he also has a periodicity constraint. Without a periodicity constraint it is easy to manufacture non-zero counterexamples. One is a linear function in the space and time coordinates

This is a valid scalar potential provided that the wave equation for the vector potential is also a solution. We can however, force by making the transformation , which in non-covariant notation is

If the transformed field can be forced to zero, then the complexity of the associated Maxwell equations are reduced. In particular, antidifferentiation of , yields

Dropping primes, the transformed Maxwell equations now take the form

There are two classes of solutions that stand out for these equations. If the vector potential is constant in time , Maxwell’s equations are reduced to the single equation

Observe that a gradient can be factored out of this equation

The solutions are then those s that satisfy both

In particular any non-time dependent potential with constant curl provides a solution to Maxwell’s equations. There may be other solutions to (3.19) too that are more general. Returning to (3.17) a second way to satisfy these equations stands out. Instead of requiring of constant curl, constant divergence with respect to the time partial eliminates (3.17). The simplest resulting equations are those for which the divergence is a constant in time and space (such as zero). The solution set are then spanned by the vectors for which

Any that both has constant divergence and satisfies the wave equation will via (2.7) then produce a solution to Maxwell’s equation.

Next in the order of complexity is consideration of the case (3.22). Here we also have , eliminated by gauge transformation, and are looking for solutions with the constraint

How can this constraint be enforced? The only obvious way is a requirement for to be zero for all , meaning that our to be determined Fourier transform coefficients are required to be perpendicular to the wave number vector parameters at all times.

The remainder of Maxwell’s equations, (3.23) impose the addition constraint on the Fourier transform

For zero equality for all it appears that we require the Fourier transforms to be harmonic in time

This has the familiar exponential solutions

also subject to a requirement that . Our field, where the are to be determined by initial time conditions, is by (2.7) of the form

Since , we have . This allows for factoring out of . The structure of the solution is not changed by incorporating the factors into , leaving the field having the general form

The original meaning of as Fourier transforms of the vector potential is obscured by the tidy up change to absorb , but the geometry of the solution is clearer this way.

It is also particularly straightforward to confirm that separately for either half of (4.34).

Case III. Non-zero scalar potential. No gauge transformation.

Now lets work from (3.11). In particular, a divergence operation can be factored from (3.11), for

Right off the top, there is a requirement for

In terms of the Fourier transforms this is

Are there any ways for this to equal a constant for all without requiring that constant to be zero? Assuming no for now, and that this constant must be zero, this implies a coupling between the and Fourier transforms of the form

A secondary implication is that or else is not a scalar. We had a transverse solution by requiring via gauge transformation that , and here we have instead the vector potential in the propagation direction.

A secondary confirmation that this is a required coupling between the scalar and vector potential can be had by evaluating the divergence equation of (4.35)

Rearranging this also produces (4.38). We want to now substitute this relationship into (3.12).

Starting with just the part we have

Taking the gradient of this brings down a factor of for

(3.12) in its entirety is now

This isn’t terribly pleasant looking. Perhaps going the other direction. We could write

so that

Note that the gradients here operate on everything to the right, including and especially the exponential. Each application of the gradient brings down an additional factor, and we have

This is identically zero, so we see that this second equation provides no additional information. That is somewhat surprising since there is not a whole lot of constraints supplied by the first equation. The function can be anything. Understanding of this curiosity comes from computation of the Faraday bivector itself. From (2.7), that is

All terms cancel, so we see that a non-zero leads to , as was the case when considering (4.24) (a case that also resulted in ).

Can this Fourier representation lead to a non-transverse solution to Maxwell’s equation? If so, it is not obvious how.

The energy momentum tensor

The energy momentum tensor is then

Observing that commutes with spatial bivectors and anticommutes with spatial vectors, and writing , the tensor splits neatly into scalar and spatial vector components

In particular for , we have

Integrating this over all space and identification of the delta function

reduces the tensor to a single integral in the continuous angular wave number space of .

Or,

Multiplying out (5.53) yields for

Recall that the only non-trivial solution we found for the assumed Fourier transform representation of was for , . Thus we have for the energy density integrated over all space, just

Observe that we have the structure of a Harmonic oscillator for the energy of the radiation system. What is the canonical momentum for this system? Will it correspond to the Poynting vector, integrated over all space?

Let’s reduce the vector component of (5.53), after first imposing the , and conditions used to above for our harmonic oscillator form energy relationship. This is

This is just

Recall that the Fourier transforms for the transverse propagation case had the form , where the minus generated the advanced wave, and the plus the receding wave. With substitution of the vector potential for the advanced wave into the energy and momentum results of (5.55) and (5.56) respectively, we have

After a somewhat circuitous route, this has the relativistic symmetry that is expected. In particular the for the complete tensor we have after integration over all space

The receding wave solution would give the same result, but directed as instead.

Observe that we also have the four divergence conservation statement that is expected

This follows trivially since both the derivatives are zero. If the integration region was to be more specific instead of a relationship, we’d have the power flux equal in magnitude to the momentum change through a bounding surface. For a more general surface the time and spatial dependencies shouldn’t necessarily vanish, but we should still have this radiation energy momentum conservation.

A complex (phasor) representation is implied, so taking real parts when all is said and done is required of the fields. For the energy momentum tensor the Geometric Algebra form, modified for complex fields, is used

The assumed four vector potential will be written

Subject to the requirement that is a solution of Maxwell’s equation

To avoid latex hell, no special notation will be used for the Fourier coefficients,

When convenient and unambiguous, this dependence will be implied.

Having picked a time and space representation for the field, it will be natural to express both the four potential and the gradient as scalar plus spatial vector, instead of using the Dirac basis. For the gradient this is

and for the four potential (or the Fourier transform functions), this is

Setup

The field bivector is required for the energy momentum tensor. This is

This last term is a spatial curl and the field is then

Applied to the Fourier representation this is

The energy momentum tensor is then

The tensor integrated over all space. Energy and momentum?

Integrating this over all space and identification of the delta function

reduces the tensor to a single integral in the continuous angular wave number space of .

Observing that commutes with spatial bivectors and anticommutes with spatial vectors, and writing , one has

The scalar and spatial vector grade selection operator has been added for convenience and does not change the result since those are necessarily the only grades anyhow. The post multiplication by the observer frame time basis vector serves to separate the energy and momentum like components of the tensor nicely into scalar and vector aspects. In particular for , one could write

If these are correctly identified with energy and momentum then it also ought to be true that we have the conservation relationship

However, multiplying out (3.12) yields for

The vector component takes a bit more work to reduce

Canceling and regrouping leaves

This has no explicit dependence, so the conservation relation (3.14) is violated unless . There is no reason to assume that will be the case. In the discrete Fourier series treatment, a gauge transformation allowed for elimination of , and this implied or constant. We will probably have a similar result here, eliminating most of the terms in (3.15) and (3.16). Except for the constant solution of the field equations there is no obvious way that such a simplified energy expression will have zero derivative.

A more reasonable conclusion is that this approach is flawed. We ought to be looking at the divergence relation as a starting point, and instead of integrating over all space, instead employing Gauss’s theorem to convert the divergence integral into a surface integral. Without math, the conservation relationship probably ought to be expressed as energy change in a volume is matched by the momentum change through the surface. However, without an integral over all space, we do not get the nice delta function cancellation observed above. How to proceed is not immediately clear. Stepping back to review applications of Gauss’s theorem is probably a good first step.

This is a doubly complex representation, with the four vector pseudoscalar acting as a non-commutatitive imaginary, as well as real and imaginary parts for the electric and magnetic field vectors. We take the real part (not the scalar part) of any bivector solution of Maxwell’s equation as the actual solution, but allow ourself the freedom to work with the complex phasor representation when convenient. In these phasor vectors, the imaginary , as in , is a commuting imaginary, commuting with all the multivector elements in the algebra.

The real valued, four vector, energy momentum tensor was found to be

To supply some context that gives meaning to this tensor the associated conservation relationship was found to be

and in particular for , this four vector divergence takes the form

relating the energy term and the Poynting spatial vector with the current density and electric field product that constitutes the energy portion of the Lorentz force density.

Let’s apply this to calculating the energy associated with the field that is periodic within a rectangular prism as done by Bohm in [2]. We do not necessarily need the Geometric Algebra formalism for this calculation, but this will be a fun way to attempt it.

Setup

Let’s assume a Fourier representation for the four vector potential for the field . That is

where summation is over all angular wave number triplets . The Fourier coefficients are allowed to be complex valued, as is the resulting four vector , and the associated bivector field .

Fourier inversion, with , follows from

but only this orthogonality relationship and not the Fourier coefficients themselves

will be of interest here. Evaluating the curl for this potential yields

Since the four vector potential has been expressed using an explicit split into time and space components it will be natural to re express the bivector field in terms of scalar and (spatial) vector potentials, with the Fourier coefficients. Writing for the spatial basis vectors, , and , this is

The Faraday bivector field is then

This is now enough to express the energy momentum tensor

It will be more convenient to work with a scalar plus bivector (spatial vector) form of this tensor, and right multiplication by produces such a split

The primary object of this treatment will be consideration of the components of the tensor, which provide a split into energy density , and Poynting vector (momentum density) .

Our first step is to integrate (12) over the volume . This integration and the orthogonality relationship (6), removes the exponentials, leaving

Because commutes with the spatial bivectors, and anticommutes with the spatial vectors, the remainder of the Dirac basis vectors in these expressions can be eliminated

Expanding the energy momentum tensor components.

Energy

In (15) only the bivector-bivector and vector-vector products produce any scalar grades. Except for the bivector product this can be done by inspection. For that part we utilize the identity

This leaves for the energy in the volume

We are left with a completely real expression, and one without any explicit Geometric Algebra. This does not look like the Harmonic oscillator Hamiltonian that was expected. A gauge transformation to eliminate and an observation about when equals zero will give us that, but first lets get the mechanical jobs done, and reduce the products for the field momentum.

Momentum

Now move on to (16). For the factors other than only the vector-bivector products can contribute to the scalar product. We have two such products, one of the form

and the other

The momentum in this volume follows by computation of

All the products are paired in nice conjugates, taking real parts, and premultiplication with gives the desired result. Observe that two of these terms cancel, and another two have no real part. Those last are

Taking the real part of this pure imaginary is zero, leaving just

I am not sure why exactly, but I actually expected a term with , quadratic in the vector potential. Is there a mistake above?

Gauge transformation to simplify the Hamiltonian.

In (20) something that looked like the Harmonic oscillator was expected. On the surface this does not appear to be such a beast. Exploitation of gauge freedom is required to make the simplification that puts things into the Harmonic oscillator form.

If we are to change our four vector potential , then Maxwell’s equation takes the form

which is unchanged by the addition of the gradient to any original potential solution to the equation. In coordinates this is a transformation of the form

and we can use this to force any one of the potential coordinates to zero. For this problem, it appears that it is desirable to seek a such that . That is

Or,

With such a transformation, the and cross term in the Hamiltonian (20) vanishes, as does the term in the four vector square of the last term, leaving just

Additionally, wedging (5) with now does not loose any information so our potential Fourier series is reduced to just

The desired harmonic oscillator form would be had in (26) if it were not for the term. Does that vanish? Returning to Maxwell’s equation should answer that question, but first it has to be expressed in terms of the vector potential. While , the lack of an component means that this can be inverted as

The gradient can also be factored scalar and spatial vector components

So, with this gauge choice the bivector field is

From the left the gradient action on is

and from the right

Taking the difference we have

Which is just

For this vacuum case, premultiplication of Maxwell’s equation by gives

The spatial bivector and trivector grades are all zero. Equating the remaining scalar and vector components to zero separately yields a pair of equations in

If the divergence of the vector potential is constant we have just a wave equation. Let’s see what that divergence is with the assumed Fourier representation

Since , there are two ways for . For each there must be a requirement for either or . The constant solution to the first equation appears to represent a standing spatial wave with no time dependence. Is that of any interest?

The more interesting seeming case is where we have some non-static time varying state. In this case, if , the second of these Maxwell’s equations is just the vector potential wave equation, since the divergence is zero. That is

Solving this isn’t really what is of interest, since the objective was just to determine if the divergence could be assumed to be zero. This shows then, that if the transverse solution to Maxwell’s equation is picked, the Hamiltonian for this field, with this gauge choice, becomes

How does the gauge choice alter the Poynting vector? From (21), all the dependence in that integrated momentum density is lost

The solutions to Maxwell’s equation are seen to result in zero momentum for this infinite periodic field. My expectation was something of the form , so intuition is either failing me, or my math is failing me, or this contrived periodic field solution leads to trouble.

Conclusions and followup.

The objective was met, a reproduction of Bohm’s Harmonic oscillator result using a complex exponential Fourier series instead of separate sine and cosines.

The reason for Bohm’s choice to fix zero divergence as the gauge choice upfront is now clear. That automatically cuts complexity from the results. Figuring out how to work this problem with complex valued potentials and also using the Geometric Algebra formulation probably also made the work a bit more difficult since blundering through both simultaneously was required instead of just one at a time.

This was an interesting exercise though, since doing it this way I am able to understand all the intermediate steps. Bohm employed some subtler argumentation to eliminate the scalar potential upfront, and I have to admit I did not follow his logic, whereas blindly following where the math leads me all makes sense.

As a bit of followup, I’d like to consider the constant case in more detail, and any implications of the freedom to pick .

The general calculation of for the assumed Fourier solution should be possible too, but was not attempted. Doing that general calculation with a four dimensional Fourier series is likely tidier than working with scalar and spatial variables as done here.

Now that the math is out of the way (except possibly for the momentum which doesn’t seem right), some discussion of implications and applications is also in order. My preference is to let the math sink-in a bit first and mull over the momentum issues at leisure.

Motivation.

We now know how to formulate the energy momentum tensor for complex vector fields (ie. phasors) in the Geometric Algebra formalism. To recap, for the field , where and may be complex vectors we have for Maxwell’s equation

This is a doubly complex representation, with the four vector pseudoscalar acting as a non-commutatitive imaginary, as well as real and imaginary parts for the electric and magnetic field vectors. We take the real part (not the scalar part) of any bivector solution of Maxwell’s equation as the actual solution, but allow ourself the freedom to work with the complex phasor representation when convenient. In these phasor vectors, the imaginary , as in , is a commuting imaginary, commuting with all the multivector elements in the algebra.

The real valued, four vector, energy momentum tensor was found to be

To supply some context that gives meaning to this tensor the associated conservation relationship was found to be

and in particular for , this four vector divergence takes the form

relating the energy term and the Poynting spatial vector with the current density and electric field product that constitutes the energy portion of the Lorentz force density.

Let’s apply this to calculating the energy associated with the field that is periodic within a rectangular prism as done by Bohm in [1]. We do not necessarily need the Geometric Algebra formalism for this calculation, but this will be a fun way to attempt it.

Setup

Let’s assume a Fourier representation for the four vector potential for the field . That is

where summation is over all wave number triplets . The Fourier coefficients are allowed to be complex valued, as is the resulting four vector , and the associated bivector field .

Fourier inversion follows from

but only this orthogonality relationship and not the Fourier coefficients themselves

will be of interest here. Evaluating the curl for this potential yields

We can now form the energy density

With implied summation over all repeated integer indexes (even without matching uppers and lowers), this is

The grade selection used here doesn’t change the result since we already have a scalar, but will just make it convenient to filter out any higher order products that will cancel anyways. Integrating over the volume element and taking advantage of the orthogonality relationship (6), the exponentials are removed, leaving the energy contained in the volume

First reduction of the Hamiltonian.

Let’s take the products involved in sequence one at a time, and evaluate, later adding and taking real parts if required all of

The expectation is to obtain a Hamiltonian for the field that has the structure of harmonic oscillators, where the middle two products would have to be zero or sum to zero or have real parts that sum to zero. The first is expected to contain only products of , and the last only products of .

While initially guessing that (13) and (14) may cancel, this isn’t so obviously the case. The use of cyclic permutation of multivectors within the scalar grade selection operator plus a change of dummy summation indexes in one of the two shows that this sum is of the form . This sum is intrinsically real, so we can neglect one of the two doubling the other, but we will still be required to show that the real part of either is zero.

Lets reduce these one at a time starting with (12), and write temporarily

So the first of our Hamiltonian terms is

Note that summation over is still implied here, so we’d be better off with a spatial vector representation of the Fourier coefficients . With such a notation, this contribution to the Hamiltonian is

To reduce (13) and (13), this time writing , we can start with just the scalar selection

Thus the contribution to the Hamiltonian from (13) and (13) is

Most definitively not zero in general. Our final expansion (15) is the messiest. Again with for short, the grade selection of this term in coordinates is

Expanding this out yields

The contribution to the Hamiltonian from this, with , is then

A final reassembly of the Hamiltonian from the parts (17) and (18) and (21) is then

This is finally reduced to a completely real expression, and one without any explicit Geometric Algebra. All the four vector Fourier vector potentials written out explicitly in terms of the spacetime split , which is natural since an explicit time and space split was the starting point.

Gauge transformation to simplify the Hamiltonian.

While (22) has considerably simpler form than (11), what was expected, was something that looked like the Harmonic oscillator. On the surface this does not appear to be such a beast. Exploitation of gauge freedom is required to make the simplification that puts things into the Harmonic oscillator form.

If we are to change our four vector potential , then Maxwell’s equation takes the form

which is unchanged by the addition of the gradient to any original potential solution to the equation. In coordinates this is a transformation of the form

and we can use this to force any one of the potential coordinates to zero. For this problem, it appears that it is desirable to seek a such that . That is

Or,

With such a transformation, the and cross term in the Hamiltonian (22) vanishes, as does the term in the four vector square of the last term, leaving just

Additionally, wedging (5) with now does not loose any information so our potential Fourier series is reduced to just

The desired harmonic oscillator form would be had in (27) if it were not for the term. Does that vanish? Returning to Maxwell’s equation should answer that question, but first it has to be expressed in terms of the vector potential. While , the lack of an component means that this can be inverted as

The gradient can also be factored scalar and spatial vector components

So, with this gauge choice the bivector field is

From the left the gradient action on is

and from the right

Taking the difference we have

Which is just

For this vacuum case, premultiplication of Maxwell’s equation by gives

The spatial bivector and trivector grades are all zero. Equating the remaining scalar and vector components to zero separately yields a pair of equations in

If the divergence of the vector potential is constant we have just a wave equation. Let’s see what that divergence is with the assumed Fourier representation

Since , there are two ways for . For each there must be a requirement for either or . The constant solution to the first equation appears to represent a standing spatial wave with no time dependence. Is that of any interest?

The more interesting seeming case is where we have some non-static time varying state. In this case, if for all the second of these Maxwell’s equations is just the vector potential wave equation, since the divergence is zero. That is

Solving this isn’t really what is of interest, since the objective was just to determine if the divergence could be assumed to be zero. This shows then, that if the transverse solution to Maxwell’s equation is picked, the Hamiltonian for this field, with this gauge choice, becomes

Conclusions and followup.

The objective was met, a reproduction of Bohm’s Harmonic oscillator result using a complex exponential Fourier series instead of separate sine and cosines.

The reason for Bohm’s choice to fix zero divergence as the gauge choice upfront is now clear. That automatically cuts complexity from the results. Figuring out how to work this problem with complex valued potentials and also using the Geometric Algebra formulation probably also made the work a bit more difficult since blundering through both simultaneously was required instead of just one at a time.

This was an interesting exercise though, since doing it this way I am able to understand all the intermediate steps. Bohm employed some subtler argumentation to eliminate the scalar potential upfront, and I have to admit I did not follow his logic, whereas blindly following where the math leads me all makes sense.

As a bit of followup, I’d like to consider the constant case, and any implications of the freedom to pick . I’d also like to construct the Poynting vector , and see what the structure of that is with this Fourier representation.

A general calculation of for an assumed Fourier solution should be possible too, but working in spatial quantities for the general case is probably torture. A four dimensional Fourier series is likely a superior option for the general case.

Motivation

The problem of a solving for the relativistically correct trajectories of classically interacting proton and electron is one that I’ve wanted to try for a while. Conceptually this is just about the simplest interaction problem in electrodynamics (other than motion of a particle in a field), but it is not obvious to me how to even set up the right equations to solve. I should have the tools now to at least write down the equations to solve, and perhaps solve them too.

Familiarity with Geometric Algebra, and the STA form of the Maxwell and Lorentz force equation will be assumed. Writing for the Faraday bivector, these equations are respectively

The possibility of self interaction will also be ignored here. From what I have read this self interaction is more complex than regular two particle interaction.

With only Coulomb interaction.

With just Coulomb (non-relativistic) interaction setup of the equations of motion for the relative vector difference between the particles is straightforward. Let’s write this out as a reference. Whatever we come up with for the relativistic case should reduce to this at small velocities.

Fixing notation, lets write the proton and electron positions respectively by and , the proton charge as , and the electron charge . For the forces we have

FIXME: picture

Subtracting the two after mass division yields the reduced mass equation for the relative motion

This is now of the same form as the classical problem of two particle gravitational interaction, with the well known conic solutions.

Using the divergence equation instead.

While use of the Coulomb force above provides the equation of motion for the relative motion of the charges, how to generalize this to the relativistic case is not entirely clear. For the relativistic case we need to consider all of Maxwell’s equations, and not just the divergence equation. Let’s back up a step and setup the problem using the divergence equation instead of Coulomb’s law. This is a bit closer to the use of all of Maxwell’s equations.

To start off we need a discrete charge expression for the charge density, and can use the delta distribution to express this.

Picking a volume element that only encloses one of the respective charges gives us the Coulomb law for the field produced by those charges as above

Here and denote the electric fields due to the proton and electron respectively. Ignoring the possibility of self interaction the Lorentz forces on the particles are

In symbols, this is

If we were to substitute back into the volume integrals we’d have

It is tempting to take the differences of these two equations so that we can write this in terms of the relative acceleration . I did just this initially, and was surprised by a mass term of the form instead of reduced mass, which cannot be right. The key to avoiding this mistake is the proper considerations of the integration volumes. Since the volumes are different and can in fact be entirely disjoint, subtracting these is not possible. For this reason we have to be especially careful if a differential form of the divergence integrals (9) were to be used, as in

The domain of applicability of these equations is no longer explicit, since each has to omit a neighborhood around the other charge. When using a delta distribution to express the point charge density it is probably best to stick with an explicit integral form.

Comparing how far we can get starting with the Gauss’s law instead of the Coulomb force, and looking forward to the relativistic case, it seems likely that solving the field equations due to the respective current densities will be the first required step. Only then can we substitute that field solution back into the Lorentz force equation to complete the search for the particle trajectories.

Relativistic interaction.

First order of business is an expression for a point charge current density four vector. Following Jackson [1], but switching to vector notation from coordinates, we can apparently employ an arbitrary parametrization for the four-vector particle trajectory , as measured in the observer frame, and write

Here is the four vector event specifying the spacetime position of the current, also as measured in the observer frame. Reparameterizating in terms of time should get us back something more familiar looking

Note that the scaling property of the delta function implies . With the split of the four-volume delta function , where , we have an explanation for why Jackson had a factor of in his representation. I initially thought this factor of was due to CGS vs SI units! One more Jackson equation decoded. We are left with the following spacetime split for a point charge current density four vector

Comparing to the continuous case where we have , it appears that this works out right. One thing worth noting is that in this time reparameterization I accidentally mixed up , the observation event coordinates of , and , the spacetime trajectory of the particle itself. Despite this, I am saved by the delta function since no contributions to the current can occur on trajectories other than , the worldline of the particle itself. So in the final result it should be correct to interpret as the spatial particle velocity as I did accidentally.

With the time reparameterization of the current density, we have for the field due to our proton and electron

How to write this in a more tidy covariant form? If we reparametrize with any of the other spatial coordinates, say we end up having to integrate the field gradient with a spacetime three form ( if parametrizing the current density with ). Since the entire equation must be zero I suppose we can just integrate that once more, and simply write

Like (7) we can pick spacetime volumes that surround just the individual particle worldlines, in which case we have a Coulomb’s law like split where the field depends on just the enclosed current. That is

Here is the field due to only the electron charge, whereas would be that part of the total field due to the proton charge.

FIXME: attempt to draw a picture (one or two spatial dimensions) to develop some comfort with tossing out a phrase like “spacetime volume surrounding a particle worldline”.

Having expressed the equation for the total field (18), we are tracking a nice parallel to the setup for the non-relativistic treatment. Next is the pair of Lorentz force equations. As in the non-relativistic setup, if we only consider the field due to the other charge we have in
in covariant Geometric Algebra form, the following pair of proper force equations in terms of the particle worldline trajectories

We have the four sets of coupled multivector equations to be solved, so the question remains how to do so. Each of the two Lorentz force equations supplies four equations with four unknowns, and the field equations are really two sets of eight equations with six unknown field variables each. Then they are all tied up together is a big coupled mess. Wow. How do we solve this?

With (19), and (21) committed to pdf at least the first goal of writing down the equations is done.

As for the actual solution. Well, that’s a problem for another night. TO BE CONTINUED (if I can figure out an attack).

In [4] many tensor quantities are not written in index form, but instead using a vector notation. In particular, the symmetric energy momentum tensor is expressed as

where the usual tensor form following by taking dot products with and substituting . The conservation equation for the canonical energy momentum tensor of (23) can be put into a similar vector form

The adjoint of the tensor can be calculated from the definition

Somewhat unintuitively, this is a function of the gradient. Playing around with factoring out the displacement vector from (26) that the energy momentum adjoint essentially provides an expansion of the gradient of the Lagrangian. To prepare, let’s introduce some helper notation

With this our Noether current equation becomes

Cyclic permutation of the vector products can be used in the scalar selection. This is a little more tractable with some helper notation for the gradients, say . Because of the operator nature of the gradient once the vector order is permuted we have to allow for the gradient to act left or right or both, so arrows are used to disambiguate this where appropriate.

This dotted with quantity is the adjoint of the canonical energy momentum tensor

This can however, be expanded further. First tackling the
bidirectional gradient vector term we can utilize the property that the reverse of a vector leaves the vector unchanged. This gives us

In the remaining term, using the Hestenes overdot notation clarify the scope of the operator, we have

The grouping of the first and third terms above simplifies nicely

Since , which is purely a trivector, the vector grade selection above is zero. This leaves the adjoint reduced to

For the remainder vector grade selection operators we have something that is of the following form

And we are finally able to put the adjoint into a form that has no remaining grade selection operators

Recapping, we have for the tensor and its adjoint

For the adjoint, since for all , we must also have , which means the adjoint of the canonical energy momentum tensor really provides not much more than a recipe for computing the Lagrangian gradient

Having seen the adjoint notation, it was natural to see what this was for a multiple scalar field variable Lagrangian, even if it is not intrinsically useful. Observe that the identity (35), obtained so laboriously, is not more than syntactic sugar for the chain rule expansion of the Lagrangian partials (plus application of the Euler-Lagrange field equations). We could obtain this directly if desired much more easily than by factoring out from .

Summing over for the gradient, this reproduces (35), with much less work

Observe that the Euler-Lagrange field equations are implied in this relationship, so perhaps it has some utility. Also note that while it is simpler to directly compute this, without having started with the canonical energy momentum tensor, we would not know how the two of these were related.

Considering an example Lagrangian we found that there was a symmetry provided we could commute the variational derivative with the gradient

What this really means is not clear in general and a better answer to the existence question for incremental translation can be had by considering the transformation of the action directly around the stationary fields.

Without really any loss of generality we can consider an action with a four dimensional spacetime volume element, and apply the incremental translation operator to this

For the first term we have , but this integral is our stationary action. The remainder, to first order in the field variables, can then be expanded and integrated by parts

Since are constants, this is zero, so there can be no contribution to the field equations by the addition of the translation increment to the Lagrangian.

General existence of the rotational symmetry.

The previous example hints at a general method to demonstrate that the incremental Lorentz transform produces a symmetry (which was assumed). It will be sufficient to consider the variation around the stationary field variables for the change due to the action from the incremental rotation operator. That is

Performing a first order Taylor expansion of the Lagrangian around the stationary field variables we have

Doing the integration by parts we have

Since for any , there is no change to the resulting field equations due to this incremental rotation, so we have a symmetry for any Lagrangian that is first order in its derivatives.

Guts

In particular, that we can use the first order approximation of this Taylor series, applying the incremental rotation operator to transform the Lagrangian.

Suppose that we parametrize the rotation bivector using two perpendicular unit vectors , and . Here perpendicular is in the sense so that . For the bivector expressed this way our incremental rotation operator takes the form

The operator is reduced to a pair of torque-like scaled directional derivatives, and we’ve already examined the Noether currents for the translations induced by the directional derivatives. It’s not unreasonable to take exactly the same approach to consider rotation symmetries as we did for translation. We found for incremental translations

So for incremental rotations the change to the Lagrangian is

Since the choice to make and both unit vectors and perpendicular has been made, there is really no loss in generality to align these with a pair of the basis vectors, say and .

The incremental rotation operator is reduced to

Similarly the change to the Lagrangian is

Subtracting the two, essentially forming , we have

We previously wrote

for the Noether current of spacetime translation, and with that our conservation equation becomes

As is, this doesn’t really appear to say much, since we previously also found . We appear to need a way to pull the x coordinates into the derivatives to come up with a more interesting statement. A test expansion of to see what is left over compared to shows that there is in fact no difference, and we actually have the identity

This suggests that we can pull the coordinates into the derivatives of (31) as in

However, expanding this derivative shows that this is fact not the case. Instead we have

So instead of a Noether current, following the procedure used to calculate the spacetime translation current, we have only a mediocre compromise

Jackson ([4]) ends up with a similar index upper expression

and then uses a requirement for vanishing 4-divergence of this quantity

to symmetrize this tensor by subtracting off all the antisymmetric portions. The differences compared to Jackson with upper verses lower indexes are minor for we can follow the same arguments and arrive at the same sort of result as we had in (31)

The only difference is that our not-really-a-conservation equation becomes

An example of the symmetry.

While not a proof that application of the incremental rotation operator is a symmetry, an example at least provides some comfort that this is a reasonable thing to attempt. Again, let’s consider the Coulomb Lagrangian

For this we have

If the variational derivative of the incremental rotation contribution is zero, then we have a symmetry.

As found in (32), we have , so we have

for this specific Lagrangian as expected.

Note that the test expansion I used to state (32) was done using only the bivector . An expansion with shows that this is also the case in shows that this is true more generally. Specifically, this expansion gives

(since the metric tensor is symmetric).

Loosely speaking, the geometric reason for this is that takes its maximum (or minimum) when is colinear with and is zero when is perpendicular to . The vector is a combined projection and 90 degree rotation in the plane of the bivector, and the divergence is left with no colinear components to operate on.

While this commutation of the with the divergence operator didn’t help with finding the Noether current, it does at least show that we have a symmetry. Demonstrating the invariance for the general Lagrangian (at least the single field variable case) likely follows the same procedure as in this specific example above.

to alter a Lagrangian density. In particular, that we can use the first order approximation of this Taylor series, applying the incremental rotation operator to transform the Lagrangian.

Suppose that we parametrize the rotation bivector using two perpendicular unit vectors , and . Here perpendicular is in the sense so that . For the bivector expressed this way our incremental rotation operator takes the form

The operator is reduced to a pair of torque-like scaled directional derivatives, and we’ve already examined the Noether currents for the translations induced by the directional derivatives. It’s not unreasonable to take exactly the same approach to consider rotation symmetries as we did for translation. We found for incremental translations

So for incremental rotations the change to the Lagrangian is

Since the choice to make and both unit vectors and perpendicular has been made, there is really no loss in generality to align these with a pair of the basis vectors, say and .

The incremental rotation operator is reduced to

Similarly the change to the Lagrangian is

Subtracting the two, essentially forming , we have

We previously wrote

for the Noether current of spacetime translation, and with that our conservation equation becomes

As is, this doesn’t really appear to say much, since we previously also found . We appear to need a way to pull the x coordinates into the derivatives to come up with a more interesting statement. A test expansion of to see what is left over compared to shows that there is in fact no difference, and we actually have the identity

The geometric reason for this is that takes its maximum (or minimum) when is colinear with and is zero when is perpendicular to . The vector is a combined projection and 90 degree rotation in the plane of the bivector, and the divergence is left with no colinear components to operate on.

FIXME: bother showing this explicitly?

The end result is that we should be able to bring the coordinates into the derivatives of (31) provided both are brought in. That gives us a more interesting conservation statement, something that has the looks of field angular momentum

The conservation identity could be summarized using

FIXME: Jackson ([4]) states a similar index upper expression

should try to show that these are identical or understand the difference.